3.65.84 \(\int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4 \log ^2(16)} \, dx\) [6484]

Optimal. Leaf size=19 \[ \frac {\left (-x+\frac {4}{x \log (16)}\right )^2}{x} \]

[Out]

(1/x/ln(2)-x)^2/x

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 14} \begin {gather*} \frac {16}{x^3 \log ^2(16)}+x-\frac {8}{x \log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48 + 8*x^2*Log[16] + x^4*Log[16]^2)/(x^4*Log[16]^2),x]

[Out]

x + 16/(x^3*Log[16]^2) - 8/(x*Log[16])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-48+8 x^2 \log (16)+x^4 \log ^2(16)}{x^4} \, dx}{\log ^2(16)}\\ &=\frac {\int \left (-\frac {48}{x^4}+\frac {8 \log (16)}{x^2}+\log ^2(16)\right ) \, dx}{\log ^2(16)}\\ &=x+\frac {16}{x^3 \log ^2(16)}-\frac {8}{x \log (16)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 20, normalized size = 1.05 \begin {gather*} x+\frac {16}{x^3 \log ^2(16)}-\frac {8}{x \log (16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48 + 8*x^2*Log[16] + x^4*Log[16]^2)/(x^4*Log[16]^2),x]

[Out]

x + 16/(x^3*Log[16]^2) - 8/(x*Log[16])

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Maple [A]
time = 0.61, size = 23, normalized size = 1.21

method	result	size
risch	\(x +\frac {-32 x^{2} \ln \left (2\right )+16}{16 \ln \left (2\right )^{2} x^{3}}\)	\(21\)
default	\(\frac {x \ln \left (2\right )^{2}+\frac {1}{x^{3}}-\frac {2 \ln \left (2\right )}{x}}{\ln \left (2\right )^{2}}\)	\(23\)
norman	\(\frac {\frac {1}{\ln \left (2\right )}+x^{4} \ln \left (2\right )-2 x^{2}}{x^{3} \ln \left (2\right )}\)	\(25\)
gosper	\(\frac {x^{4} \ln \left (2\right )^{2}+1-2 x^{2} \ln \left (2\right )}{x^{3} \ln \left (2\right )^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(16*x^4*ln(2)^2+32*x^2*ln(2)-48)/x^4/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

1/ln(2)^2*(x*ln(2)^2+1/x^3-2*ln(2)/x)

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Maxima [A]
time = 0.29, size = 26, normalized size = 1.37 \begin {gather*} \frac {x \log \left (2\right )^{2} - \frac {2 \, x^{2} \log \left (2\right ) - 1}{x^{3}}}{\log \left (2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="maxima")

[Out]

(x*log(2)^2 - (2*x^2*log(2) - 1)/x^3)/log(2)^2

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Fricas [A]
time = 0.35, size = 25, normalized size = 1.32 \begin {gather*} \frac {x^{4} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (2\right ) + 1}{x^{3} \log \left (2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="fricas")

[Out]

(x^4*log(2)^2 - 2*x^2*log(2) + 1)/(x^3*log(2)^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).
time = 0.04, size = 24, normalized size = 1.26 \begin {gather*} \frac {x \log {\left (2 \right )}^{2} + \frac {- 2 x^{2} \log {\left (2 \right )} + 1}{x^{3}}}{\log {\left (2 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*x**4*ln(2)**2+32*x**2*ln(2)-48)/x**4/ln(2)**2,x)

[Out]

(x*log(2)**2 + (-2*x**2*log(2) + 1)/x**3)/log(2)**2

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Giac [A]
time = 0.38, size = 26, normalized size = 1.37 \begin {gather*} \frac {x \log \left (2\right )^{2} - \frac {2 \, x^{2} \log \left (2\right ) - 1}{x^{3}}}{\log \left (2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(16*x^4*log(2)^2+32*x^2*log(2)-48)/x^4/log(2)^2,x, algorithm="giac")

[Out]

(x*log(2)^2 - (2*x^2*log(2) - 1)/x^3)/log(2)^2

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Mupad [B]
time = 0.05, size = 18, normalized size = 0.95 \begin {gather*} \frac {{\left (x^2\,\ln \left (2\right )-1\right )}^2}{x^3\,{\ln \left (2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*log(2)^2 + 2*x^2*log(2) - 3)/(x^4*log(2)^2),x)

[Out]

(x^2*log(2) - 1)^2/(x^3*log(2)^2)

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